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Question Number 206160 by MATHEMATICSAM last updated on 08/Apr/24

If x and y are real numbers then is it possible that secθ = ((2xy)/(x^2 + y^2 )) ?

$$\mathrm{If}\:{x}\:\mathrm{and}\:{y}\:\mathrm{are}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{then}\:\mathrm{is}\:\mathrm{it} \\ $$$$\mathrm{possible}\:\mathrm{that}\:\mathrm{sec}\theta\:=\:\frac{\mathrm{2}{xy}}{{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} }\:? \\ $$

Commented by mr W last updated on 08/Apr/24

only if x=±y.

$${only}\:{if}\:{x}=\pm{y}. \\ $$

Answered by mahdipoor last updated on 08/Apr/24

sec a=((2xy)/(x^2 +y^2 ))∈]−1,1[⇒ ((2xy)/(x^2 +y^2 ))≤−1⇒2xy≤−x^2 −y^2 ⇒(x+y)^2 ≤0 ⇒x=−y ((2xy)/(x^2 +y^2 ))≥1⇒2xy≥x^2 +y^2 ⇒0≥(x−y)^2 ⇒x=y ⇒⇒x=±y

$$\left.{sec}\:{a}=\frac{\mathrm{2}{xy}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\in\right]−\mathrm{1},\mathrm{1}\left[\Rightarrow\right. \\ $$$$\frac{\mathrm{2}{xy}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\leqslant−\mathrm{1}\Rightarrow\mathrm{2}{xy}\leqslant−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \Rightarrow\left({x}+{y}\right)^{\mathrm{2}} \leqslant\mathrm{0} \\ $$$$\Rightarrow{x}=−{y} \\ $$$$\frac{\mathrm{2}{xy}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\geqslant\mathrm{1}\Rightarrow\mathrm{2}{xy}\geqslant{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \Rightarrow\mathrm{0}\geqslant\left({x}−{y}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{x}={y} \\ $$$$\Rightarrow\Rightarrow{x}=\pm{y} \\ $$

Commented by Frix last updated on 08/Apr/24

sec α =(1/(cos α)) ⇒ sec α ∈(−∞, −1]∪[1, ∞)

$$\mathrm{sec}\:\alpha\:=\frac{\mathrm{1}}{\mathrm{cos}\:\alpha}\:\Rightarrow\:\mathrm{sec}\:\alpha\:\in\left(−\infty,\:−\mathrm{1}\right]\cup\left[\mathrm{1},\:\infty\right) \\ $$

Answered by Frix last updated on 08/Apr/24

sec θ =(1/(cos θ)) cos θ =((x^2 +y^2 )/(2xy))=c x^2 +y^2 =2cxy y^2 −2cxy+x^2 =0 y=(c±(√(c^2 −1)))x y=(cos θ ±i sin θ)x y∈R ⇒ θ=nπ ⇒ cos θ =±1 ⇒ y=±x

$$\mathrm{sec}\:\theta\:=\frac{\mathrm{1}}{\mathrm{cos}\:\theta} \\ $$$$\mathrm{cos}\:\theta\:=\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}{xy}}={c} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2}{cxy} \\ $$$${y}^{\mathrm{2}} −\mathrm{2}{cxy}+{x}^{\mathrm{2}} =\mathrm{0} \\ $$$${y}=\left({c}\pm\sqrt{{c}^{\mathrm{2}} −\mathrm{1}}\right){x} \\ $$$${y}=\left(\mathrm{cos}\:\theta\:\pm\mathrm{i}\:\mathrm{sin}\:\theta\right){x} \\ $$$${y}\in\mathbb{R}\:\Rightarrow\:\theta={n}\pi\:\Rightarrow\:\mathrm{cos}\:\theta\:=\pm\mathrm{1}\:\Rightarrow\:{y}=\pm{x} \\ $$

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